Question

${\int }_{\log \frac{1}{2}}^{\log 2}\sin (\frac{e^x-1}{e^x+1})dx$ is equal to

• A. $\cos \frac{1}{3}$
• B. 0
• C. $2\cos 2$
• D. none of these

Answer 1

Answer: (B)

Let $f\left(x\right)=sin\left(\frac{e^x-1}{e^x+1}\right)$
$\therefore f\left(-x\right)=sin\left(\frac{e^{-x}-1}{e^{-x}+1}\right)=sin\left(\frac{\frac{1}{e^x}-1}{\frac{1}{e^x}+1}\right)$
$=sin\left(\frac{1-e^x}{1+e^x}\right)=-sin\left(\frac{e^x-1}{e^x+1}\right)=-f\left(x\right)$
$\therefore f\left(x\right)$ is an odd function.
$\therefore$ given integral $={\int }_{-log2}^{log2}sin\left(\frac{e^x-1}{e^x+1}\right)dx=0$